21 Mark for Review If 3x^(2)-18 x-15=0, what is the value of x^(2)-6x ?

Factor the quadratic equation: Given the equation 3x^2 - 18x - 15 = 0, we want to find the value of x^2 - 6x. To do this, we can first solve for x by factoring or using the quadratic formula.Factor by grouping: Let's try to factor the quadratic equation. We look for two numbers that multiply to (3)(-15) = -45 and add up to -18. The numbers that satisfy these conditions are -15 and -3.Find possible solutions for x: We can write the quadratic equation as 3x^2 - 15x - 3x - 15 = 0. Now, we can factor by grouping.Substitute x values: Grouping the terms, we get (3x^2 - 15x) - (3x + 15) = 0. Factoring out the common factors, we have 3x(x - 5) - 3(x + 5) = 0.Determine single value for x^2 - 6x: Now, we can factor out a 3 from both groups, getting 3(x - 5)(x + 5) = 0.Setting each factor equal to zero gives us the possible solutions for x: x - 5 = 0 or x + 5 = 0. Therefore, x = 5 or x = -5.Now that we have the values of x, we can substitute them into the expression x^2 - 6x to find the corresponding values.First, let's substitute x = 5 into x^2 - 6x: (5)^2 - 6(5) = 25 - 30 = -5.Next, let's substitute x = -5 into x^2 - 6x: (-5)^2 - 6(-5) = 25 + 30 = 55.We have two possible values for x^2 - 6x, which are -5 and 55. However, since the original equation is a quadratic, it should have a single value for x^2 - 6x that satisfies the equation for both roots.To find the single value that works for both roots, we can use the original equation 3x^2 - 18x - 15 = 0 and divide through by 3 to simplify it to x^2 - 6x - 5 = 0.Now, we can see that x^2 - 6x is equal to 5 because x^2 - 6x - 5 = 0 implies x^2 - 6x = 5.

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Algebra 2

Evaluate expression when a complex numbers and a variable term is given

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Q. If

3x^{2}-18x-15=0

, what is the value of

x^{2}-6x

Factor the quadratic equation: Given the equation $3x^2 - 18x - 15 = 0$ , we want to find the value of $x^2 - 6x$ . To do this, we can first solve for $x$ by factoring or using the quadratic formula.
Factor by grouping: Let's try to factor the quadratic equation. We look for two numbers that multiply to $(3)(-15) = -45$ and add up to $-18$ . The numbers that satisfy these conditions are $-15$ and $-3$ .
Find possible solutions for x: We can write the quadratic equation as $3x^2 - 15x - 3x - 15 = 0$ . Now, we can factor by grouping.
Substitute x values: Grouping the terms, we get $(3x^2 - 15x) - (3x + 15) = 0$ . Factoring out the common factors, we have $3x(x - 5) - 3(x + 5) = 0$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .Next, let's substitute $x = -5$ into $x^2 - 6x$ : $3$ $5$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .Next, let's substitute $x = -5$ into $x^2 - 6x$ : $3$ $5$ .We have two possible values for $x^2 - 6x$ , which are $3$ $7$ and $3$ $8$ . However, since the original equation is a quadratic, it should have a single value for $x^2 - 6x$ that satisfies the equation for both roots.
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .Next, let's substitute $x = -5$ into $x^2 - 6x$ : $3$ $5$ .We have two possible values for $x^2 - 6x$ , which are $3$ $7$ and $3$ $8$ . However, since the original equation is a quadratic, it should have a single value for $x^2 - 6x$ that satisfies the equation for both roots.To find the single value that works for both roots, we can use the original equation $3(x - 5)(x + 5) = 0$ $0$ and divide through by $3$ to simplify it to $3(x - 5)(x + 5) = 0$ $2$ .
Determine single value for $x^2 - 6x$ : Now, we can factor out a $3$ from both groups, getting $3(x - 5)(x + 5) = 0$ .Setting each factor equal to zero gives us the possible solutions for $x$ : $x - 5 = 0$ or $x + 5 = 0$ . Therefore, $x = 5$ or $x = -5$ .Now that we have the values of $x$ , we can substitute them into the expression $x^2 - 6x$ to find the corresponding values.First, let's substitute $x = 5$ into $x^2 - 6x$ : $3$ $2$ .Next, let's substitute $x = -5$ into $x^2 - 6x$ : $3$ $5$ .We have two possible values for $x^2 - 6x$ , which are $3$ $7$ and $3$ $8$ . However, since the original equation is a quadratic, it should have a single value for $x^2 - 6x$ that satisfies the equation for both roots.To find the single value that works for both roots, we can use the original equation $3(x - 5)(x + 5) = 0$ $0$ and divide through by $3$ to simplify it to $3(x - 5)(x + 5) = 0$ $2$ .Now, we can see that $x^2 - 6x$ is equal to $3(x - 5)(x + 5) = 0$ $4$ because $3(x - 5)(x + 5) = 0$ $2$ implies $3(x - 5)(x + 5) = 0$ $6$ .

If $3x^{2}-18x-15=0$ , what is the value of $x^{2}-6x$ ?

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If $3x^{2}-18x-15=0$ , what is the value of $x^{2}-6x$ ?